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18 Aug 2021, 08:00
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E
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Difficulty:
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55% (02:15) correct
45%
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wrong
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In rectangle with side length of 10 cm, two identical semicircles are drawn with their diameters coinciding with the opposite sides of the rectangle. Then a circle is drawn through the points A, B, C and D, where A and C are the intersection points of the semicircles. What is the area of the green region?
A. \(10.5\pi - 25\)
B. \(11\pi - 25\)
C. \(11.5\pi - 25\)
D. \(12\pi - 25\)
E. \(12.5\pi - 25\)
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18 Aug 2021, 09:10
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B and D bisects the opposite sides of the rectangle and are thus the centers of the two identical semicircles, and the radius of them is 5cm.
Connecting A, B, C, D and we have a square. This is because <ADC and <ABC are both right angles (because AC is the diameter), and AB = BC = CD = DA = 5cm (all 4 are radii of the two identical semicircles).
The area for the green area can be calculated as the areas of sector DAC + BAC - the area of the square ABCD.
The areas of sector DAC and BAC are the same and = 90 degrees / 360 degrees * pie * 5^2 = 25/4 * pie. The area of the square = 5^2 = 25.
Therefore the area of the green area = 2 * 25/4 * pie - 25 = 12.5*pie - 25. So E).
Connecting A, B, C, D and we have a square. This is because <ADC and <ABC are both right angles (because AC is the diameter), and AB = BC = CD = DA = 5cm (all 4 are radii of the two identical semicircles).
The area for the green area can be calculated as the areas of sector DAC + BAC - the area of the square ABCD.
The areas of sector DAC and BAC are the same and = 90 degrees / 360 degrees * pie * 5^2 = 25/4 * pie. The area of the square = 5^2 = 25.
Therefore the area of the green area = 2 * 25/4 * pie - 25 = 12.5*pie - 25. So E).
19 Nov 2021, 02:46
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Official Solution:
In rectangle with side length of 10 cm, two identical semicircles are drawn with their diameters coinciding with the opposite sides of the rectangle. Then a circle is drawn through the points A, B, C and D, where A and C are the intersection points of the semicircles. What is the area of the green region?
A. \(10.5\pi - 25\)
B. \(11\pi - 25\)
C. \(11.5\pi - 25\)
D. \(12\pi - 25\)
E. \(12.5\pi - 25\)
Consider triangle ACD below:
Recall that A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle's side, then that triangle is a right triangle. Since AC is the diameter of inscribed small circle then ACD is a right triangle and D is a right angle. Next, DA and DC are radii of the bottom semicircle, so \(DA=DC=\frac{10}{2}=5\). So, the area of triangle ADC is \(\frac{1}{2}*5*5=\frac{25}{2}\)
Now, consider sector ACD. Since D = 90°, then the area of this sector is half the area of the bottom semicircle (the area of 90° sector is \(\frac{1}{4}\) th the area of a circle, so it's half the area of the semicircle). The area of the semicircle is \(\frac{\pi r^2}{2}=\frac{25\pi}{2}\) and thus the area of sector ACD is \(\frac{25\pi}{4}\).
Notice that if we subtract the area of triangle ACD from the area of sector ACD, we'll get HALF the area of the green region. So, the area of the green region is:
\(2*(\frac{25\pi}{4} -\frac{25}{2})=\frac{25\pi}{2}-25\)
Answer: E
In rectangle with side length of 10 cm, two identical semicircles are drawn with their diameters coinciding with the opposite sides of the rectangle. Then a circle is drawn through the points A, B, C and D, where A and C are the intersection points of the semicircles. What is the area of the green region?
A. \(10.5\pi - 25\)
B. \(11\pi - 25\)
C. \(11.5\pi - 25\)
D. \(12\pi - 25\)
E. \(12.5\pi - 25\)
Consider triangle ACD below:
Recall that A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle's side, then that triangle is a right triangle. Since AC is the diameter of inscribed small circle then ACD is a right triangle and D is a right angle. Next, DA and DC are radii of the bottom semicircle, so \(DA=DC=\frac{10}{2}=5\). So, the area of triangle ADC is \(\frac{1}{2}*5*5=\frac{25}{2}\)
Now, consider sector ACD. Since D = 90°, then the area of this sector is half the area of the bottom semicircle (the area of 90° sector is \(\frac{1}{4}\) th the area of a circle, so it's half the area of the semicircle). The area of the semicircle is \(\frac{\pi r^2}{2}=\frac{25\pi}{2}\) and thus the area of sector ACD is \(\frac{25\pi}{4}\).
Notice that if we subtract the area of triangle ACD from the area of sector ACD, we'll get HALF the area of the green region. So, the area of the green region is:
\(2*(\frac{25\pi}{4} -\frac{25}{2})=\frac{25\pi}{2}-25\)
Answer: E
